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FIGURE 4.1 below is a block diagram of the national grid - NSC Electrical Technology Power Systems - Question 4 - 2021 - Paper 1

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FIGURE 4.1 below is a block diagram of the national grid. Answer the questions that follow. 4.1 Name the processes at A, B and C. 4.1.2 Draw the voltage waveforms ... show full transcript

Worked Solution & Example Answer:FIGURE 4.1 below is a block diagram of the national grid - NSC Electrical Technology Power Systems - Question 4 - 2021 - Paper 1

Step 1

Name the processes at A, B and C.

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Answer

The processes can be identified as follows:

  • A: Generation
  • B: Transmission
  • C: Distribution

Step 2

Draw the voltage waveforms generated at A.

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Answer

The voltage waveforms can be drawn as follows:

  1. Three sinusoidal waveforms with equal amplitude and a phase difference of 120° between each.
  2. The waveforms will be placed on the same time axis, showing the peaks and zero crossings distinctly.

Step 3

Explain why Transformer 1 is a step-up transformer.

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Answer

Transformer 1 is a step-up transformer because it increases the voltage received from the generation process to a higher level for transmission. This is done to reduce current flow in the lines, thereby minimizing I²R losses during transmission.

Step 4

Determine the type of connection used on the secondary windings of Transformer 3 and motivate your answer.

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Answer

The type of connection used is a star connection. This configuration allows for both three-phase and single-phase equipment to be served, making it versatile for varying load types.

Step 5

Calculate the phase current.

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Answer

The phase current can be calculated using the formula: Iph=IL31538.66AI_{ph} = \frac{I_L}{\sqrt{3}} \Rightarrow \frac{15}{\sqrt{3}} \approx 8.66 A

Step 6

Calculate the apparent power.

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Answer

The apparent power (S) can be calculated using: S=3×VL×IL3×380×159872.69VA9.87kVAS = \sqrt{3} \times V_L \times I_L \Rightarrow \sqrt{3} \times 380 \times 15 \approx 9872.69 VA \approx 9.87 kVA

Step 7

Calculate the phase angle.

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Answer

The phase angle (θ) can be calculated using: θ=cos1(0.9)25.84\theta = \cos^{-1}(0.9) \approx 25.84^{\circ}

Step 8

Calculate the reactive power.

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Answer

The reactive power (Q) is calculated using: Q=3×VL×IL×sinθ3×380×15×sin(25.84)4.3kVAQ = \sqrt{3} \times V_L \times I_L \times \sin\theta \Rightarrow \sqrt{3} \times 380 \times 15 \times \sin(25.84^{\circ}) \approx 4.3 kVA

Step 9

State TWO advantages of power factor correction to the consumer.

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Answer

  1. Reduced current drawn from the supply, leading to lower electricity bills.
  2. Increased lifespan and reduced maintenance of equipment due to lower heat generation.

Step 10

Does the reading show a leading or lagging power factor?

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Answer

The meter shows a lagging power factor.

Step 11

State the cause of this type of reading.

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Answer

The connected load is predominantly inductive, which results in a lagging power factor.

Step 12

State how the meter reading could be brought closer to unity.

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Answer

By connecting capacitors in parallel with the load or using synchronous motors, we can improve the power factor and bring the meter reading closer to unity.

Step 13

Calculate the total power of the load.

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Answer

The total power of the load can be calculated as: Pt=P1+P2=250W+460W=710WP_t = P_1 + P_2 = 250 W + 460 W = 710 W

Step 14

State TWO quantities, other than power, that can be determined by using the two-wattmeter method.

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Answer

  1. Power factor.
  2. Phase angle.

Step 15

Explain why the two-wattmeter method is preferred over the three-wattmeter method.

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Answer

The two-wattmeter method is simpler and easier to connect to the lines, and it is also a more economical method of load measurement.

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